What is the concentration of the 100 g of solution having 25 g of solute?

PLEASE HELP!! I'll mark brainliest for correct answer.

Tju [1.3M]

1. Car A
2. Car C
3. Car A
4. 1500 J
5. 750 J
6. 750 J
7. Car A was had a height higher relative to the ground compared to all the other cars.
8. mgh
9. B

6 0

3 years ago

A sphere with a charge q is fixed at the bottom left corner of the right triangle shown in the figure. Points P and R are at the

Alexus [3.1K]

Answer:

the final potential energy of this system is 3U0/10

Explanation:

We are given

charge at left end and another test charge at point p

Potential energy is given by = $\frac{k*Q1*Q2 }{R}$

where k is electrostatics constant = $9 *10^9$

Q1 = first charge , Q2= test charge

R= distance between charges

potential at point p

U0 = k*Q1*Q2 /3 ⇒ kq1q2 = 3U0 ..............1

now the test charge moves to point R

using Pytahgoreou theorem

R(distance) = $\sqrt{8^2 + 6^2}$ = 10

New Potential energy

U1 = kq1*q2 / 10

substituting kq1q2 = 3U0 from 1

U1 = 3U0/10

So this is the final potential energy of this system.

5 0

2 years ago

Calculate the potential difference across a 25- resistor if a 0.3-A current is flowing through it.

alekssr [168]

Voltage = (current) x (resistance)

Voltage = (0.3 A) x (25 ohms)

Voltage = (0.3 x 25) (A-Ω)

<em>Voltage = 7.5 volts</em>

8 0

3 years ago

Read 2 more answers

For a series circuit what is the terminal voltage of a battery or power supply equal to in terms of the potential difference or

trapecia [35]

Answer: V=IR

Explanation: for a series circuit connected to a battery supply, the total emf across the circuit is given as

E = I(R + r) and by expanding, we have that E =IR + It

Where r is the internal resistance of the battery

I is the total current flowing in the circuit

R total load resistance in the circuit.

E is the total emf of the circuit.

The total emf is the sum of 2 separate voltages.

"IR" which is the terminal voltage and "Ir" which is the loss voltage.

The teenila voltage is the voltage flowing in the circuit based on the equivalent resistance of the circuit while the loss voltage is the wasted voltage based on the internal resistance of the battery source.

7 0

3 years ago

What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500

padilas [110]

Answer:

I = 1.21x10^-5 A

Explanation:

You are missing the first part of the problem. This is an example, but it will give you the idea of how to solve yours with your data.

The first part is like this:

<em>A 4.0 cm diameter parallel plate capacitor has a 0.44 m m gap. What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000 V/s?</em>

Now with this, we can solve the problem.

In order to do this, we need to use the following expression:

q = CV (1)

Where:

C: Capacitance of a parellel capacitor (in Faraday)

q: charge of plate or capacitor (In coulombs)

V: voltage in Volts.

However, we need is the current, and we have data of potential difference, so, all we have to do is divide the expression between time so:

q/t = CV/t

And the current is q/t, thus:

I = C * V/t (2)

And finally, Capacitance C with two plates of area A separated by a distance d is:

C = Eo*A/d (3)

Where:

Eo = constant equals to 8.85x10^-12 F/m.

A = Area of the plate, in this case, πr²

d = gap of the capacitor.

Let's calculate first the Capacitance using equation (3):

C = 8.85x10^-12 * π * (0.04/2)² / 0.00046 = 2.42x10^-11 F

Now, it's time to use equation (2) and solve for I:

I = 2.42x10^-11 * 500,000

I = 1.21x10^-5 A

5 0

3 years ago